$N$ real numbers such that many $9$-sets have integer sum.

Question

We have $N\geq 9$ distinct positive reals numbers in range $[0,1)$ such that whenever we pick $8$ of them, there is another one (different from the first $8$) so that their sum is an integer. What are the possible values of $N$?

Clearly $N=9$ works, just take $9$ distinct reals adding to $1$. But I have not been able to figure out other values. However, we can notice that there are at least $\frac{\binom{N}{8}}{9}$ subsets of $9$ elements which have an integer sum, and this sum must be among $\{1,2,3,4,5,6,7,8\}$. So at least one of the numbers appears at least $\frac{\binom{N}{8}}{72}$ times. I get a feeling that this is impossible for large $N$.

Answer 1

Partial result: The case $N=10$ is impossible.

Pf: Suppose we had $10$ of them, sorted as $0<a_1<\cdots <a_{10}$.

let $x=a_1+\cdots a_8$ We know that either $x+a_9$ or $x+a_{10}$ is an integer.

Remark: they can't both be integers as $0<a_{10}-a_9<1$

Case I: $x+a_9$ is an integer.

Then define $y=a_2+\cdots +a_8+a_{10}$ Which of $a_1$ and $a_9$ should we add to get an integer? It can't be $a_1$ by the remark above, so it must be $a_9$. But then $(y+a_9)-(x+a_9)=y-x=a_{10}-a_1 \in \mathbb Z$, a contradiction.

Case II: $x+a_{10}$ is an integer.

Then define $y=a_2+\cdots a_{9}$ Which of $a_1$ and $a_{10}$ should we add to get an integer? It can't be $a_1$ by the remark above, so it must be $a_{10}$. But then $(y+a_{10})-(x+a_{10})=y-x=a_{9}-a_1 \in \mathbb Z$, a contradiction.